Determining which sets are spanning sets for P3? I have a list of sets:
(a): {$1, x^2, x^2 - 2$} 
(b): {$2, x^2, x, 2x + 3$} 
(c): {$x + 2, x + 1, x^2 - 1$} 
(d): {$x + 2, x^2 -1$}
I am supposed to determine which of the following are spanning sets in $P_3$.
My attempted solution:

(A): Ax^2 + bx + c = A(1) + B(x^2) +C(x^2-2)

 Ax^2 + bx + c = (A-2C) + Bx^2 + C(x^2)

This is pretty much as far as I've got and I am not exactly sure what I'm doing or solving for. Any clarification would be wonderful. 
 A: A spanning set for this set of polynomials, $ \ ax^2 \ + \ bx \ + \ c \ , $ must be a collection of polynomials, $ \ p_i(x) \ , $ that can be combined by linear combinations to produce all possible polynomials of second-degree or lower, including one-term polynomials, such as $ \ ax^2 \ , \ bx \ , $ and $ \ c \ ; $ the "constant polynomials" must include the "zero polynomial" , $ \ y = 0 \ . $ 
What we are doing is constructing a "vector space" of polynomials*, so it must be closed under addition and scalar multiplication (thus, under linear combination) and it must include a "zero element".  There is no limit to the number of "vectors" (in this case, polynomials) that can appear in a spanning set, as long as all the polynomials in $ \ P_3 \ $ can be produced.  (For esthetic reasons -- or reasons of sanity -- we usually prefer to find "minimal" spanning sets, ones with the fewest possible elements.)
$ * $ We don't ordinarily think of polynomials as "vectors", but in this context, the arrangement of the polynomials under linear combinations has the "algebraic structure" of a vector space.  So, in recent times, we talk about vector spaces of lots of mathematical objects that don't seem like vectors...
So we need to be able to produce $ \ ax^2 \ + \ bx \ + \ c \  $ from $ \ C_1 \cdot p_1(x) \ + \  C_2 \cdot p_2(x) \ + \ \ldots \ + \ C_n \cdot p_n(x) . $  There are two choices we can eliminate immediately.  Choice (D) only permits
$$ ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( x + 2 ) \ + \ C_2 \ (x^2 - 1) \ = \ C_2x^2 \ + \ C_1 x + ( 2C_1 - C_2 ) \ , $$
which "forces" the choices $ \ C_1 = b \ \ \text{and} \ \ C_2 = a \ , $ which permits us no control at all as to what value of $ \ c = 2C_1 - C_2 \ $ we'll end up with.  Further, we can't produce a polynomial $ \ ax^2 \ $ or $ \ bx \ $ without having to have a non-zero constant term, nor can we get a "constant polynomial $ \ c \ $ without having other terms along with it.  So (D) is definitely not a spanning set.  Choice (A) is even worse because it doesn't even give us a way to produce a polynomial $ \ bx \  $ through linear combination, as none of the polynomials in the set have a linear term.
This leaves choices (B) and (C).  It turns out both of these are spanning sets, as E-theory indicates. For (C), we have
$$ ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( x + 2 ) \ + \ \ C_2 \ ( x + 1 ) \ + \ C_3 \ (x^2 - 1) $$
$$ = \ C_3x^2 \ + \ (C_1  + C_2) x + ( 2C_1 + C_2 - C_3) \ , $$
which corrects the deficiency of choice (D) by leaving us a "measure of control" over each coefficient:  $ \ C_3 \ $ must equal $ \ a \ , $ but then we can adjust $ \ C_2 \ $ in the sum $ \ C_1 + C_2 \ $ in order to match $ \ b \ , $ and in turn adjust $ \ C_1 \ $ in $ \ 2C_1 + C_2 - C_3 \ $ to give us $ \ c \ . $  This will mean that it is possible to calculate just what each unknown $ \ C_i \ $ must be in order to make any of the polynomials in $ \ P_3 \ . $  Note also that we must match three coefficients in the polynomial and we have three "basis polynomials" to manage it with; this is a "minimal spanning set".
By the same token, choice (B) gives
$$ ax^2 \ + \ bx \ + \ c \ = \ C_1 \ ( 2 ) \ + \ \ C_2 \ ( x^2 ) \ + \ C_3 \ (x) \ + \ C_4 \ (2x+3) . $$
We can see pretty readily that every power of $ \ x \ $ needed to "build" second-degree polynomials is present in the first three elements of the set, so we can just use $ \ C_1 = \frac{c}{2} \ , \ C_2 = a \ , \ \ \text{and} \ \ C_3 = b \ . $  What about $  \ (2x+3) \ $ ?  Well, it can certainly be incorporated into the calculation of the values for obtaining $ \ b \ \ \text{and} \ \ c \ , $ but we've also just seen that it isn't needful.  So we can also choose to just "turn off" its contribution by always setting $ \ C_4 = 0 \ . $  So choice (B) also gives us a spanning set for $ \ P_3 \ $ ; it just isn't minimal.
A: I found b) and c) are the spanning sets. 
For b) You only need to show that you can find reals m, n, p such that:
ax^2 + bx + c = 2m + nx^2 + px. You don't need 2x + 3 since 2x + 3 = 2x + (3/2)*2 is in the span{2,x}. So choose n = a, p = b, and m = c/2.
For c) again ax^2 + bx + c = m(x+2) + n(x+1) + p(x^2 - 1). Choose p = a, and :
m = b - a - c, and n = a + c. 
For a) x^2 + 2x + 1 is not in the span{1, x^2, x^2 - 2}.
For d) ax^2 + bx + c = m(x+2) + n(x^2 -1) ==> n = a , m = b, c = 2m - n = 2b - a. So you only need to find a quadratic polynomial like x^2 + x - 5, then it is not in the span{x+2, x^2 - 1}.
A: $P_3$ has the basis $\{1,x,x^2\}$.  Since each set $S$ is a subset of $P_3$, we have that $S$ spans $P_3$ if and only if $1$, $x$ and $x^2$ can be formed by linear combinations of the polynomials in $S$.
(a): $S=\{1, x^2, x^2 - 2\}$: can't form $x$ as a linear combination.  So $S$ does not span $P^3$.
(b): $S=\{2, x^2, x, 2x + 3\}$: We have the linear combinations $$1=\tfrac{1}{2} 2+0x^2+0x+0(2x+3);$$ $$x=0 \times 2+0x^2+1x+0(2x+3);$$ $$x^2=0 \times 2+1x^2+0x+0(2x+3),$$ so $S$ spans $P^3$.
The other two are of the same flavour: either show (a) the polynomials $1$, $x$ and $x^2$ are linear combinations of the polynomials in $S$, or (b) that one of them cannot be formed as a linear combination for some reason.
